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Answer :

Given,

The total charge transferred, Q=15.0 C

The charge of one electron is e=1.6×10⁻¹⁹ C

Thus the number of electrons in 15.0 C is given by,

[tex]N=\frac{Q}{e}[/tex]

On substituting the known values,

[tex]\begin{gathered} N=\frac{15}{1.6\times10^{-19}} \\ =9.38\times10^{19} \end{gathered}[/tex]

Therefore the number of electrons that pass through the wire is 9.38×10¹⁹

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